Logarithmic Differentiation — Question 3

Question 3 - Solution

We are given: $$f(x) = \frac{(3x + 1)^4 \cdot \ln(x)}{(x^2 - 9)^{3/2}}$$

Step 1: Take the natural log of both sides:

$$\ln f(x) = \ln \left( \frac{(3x + 1)^4 \cdot \ln(x)}{(x^2 - 9)^{3/2}} \right)$$

Apply logarithmic identities: $$\ln f(x) = 4 \ln(3x + 1) + \ln(\ln x) - \frac{3}{2} \ln(x^2 - 9)$$

Step 2: Differentiate both sides implicitly:

$$\frac{f'(x)}{f(x)} = \frac{4 \cdot 3}{3x + 1} + \frac{1}{\ln x} \cdot \frac{1}{x} - \frac{3}{2} \cdot \frac{2x}{x^2 - 9}$$

Simplify: $$\frac{f'(x)}{f(x)} = \frac{12}{3x + 1} + \frac{1}{x \ln x} - \frac{3x}{x^2 - 9}$$

Step 3: Multiply both sides by $f(x)$ to isolate $f'(x)$:

$$f'(x) = f(x) \left( \frac{12}{3x + 1} + \frac{1}{x \ln x} - \frac{3x}{x^2 - 9} \right)$$

Substitute back the original $f(x)$:

$$f'(x) = \frac{(3x + 1)^4 \cdot \ln(x)}{(x^2 - 9)^{3/2}} \left( \frac{12}{3x + 1} + \frac{1}{x \ln x} - \frac{3x}{x^2 - 9} \right)$$

$$\boxed{ f'(x) = \frac{(3x + 1)^4 \cdot \ln(x)}{(x^2 - 9)^{3/2}} \left( \frac{12}{3x + 1} + \frac{1}{x \ln x} - \frac{3x}{x^2 - 9} \right) }$$